The coefficient of linear expansion of brass is \(1.9 \times 10^{-5}\) $^{\circ} \mathrm{C}^{-1} .\( At \)20.0^{\circ} \mathrm{C},$ a hole in a sheet of brass has an area of \(1.00 \mathrm{mm}^{2} .\) How much larger is the area of the hole at $30.0^{\circ} \mathrm{C} ?(\text { Wile tutorial: loop around the equator })$

The change in temperature (\(\Delta T\)) is given by the final temperature minus the initial temperature. In this case, the initial temperature is \(20.0^{\circ} \mathrm{C}\), and the final temperature is \(30.0^{\circ} \mathrm{C}\). Therefore, \(\Delta T = 30.0 - 20.0 = 10.0^{\circ} \mathrm{C}\).

We can use the formula for linear expansion to find the change in length (\(\Delta L\)) of a side of the hole: \(\Delta L = L \times \alpha \times \Delta T\) Where \(L\) is the initial length, \(\alpha\) is the coefficient of linear expansion, and \(\Delta T\) is the change in temperature. Since we are given the initial area of the hole (\(1.00 \mathrm{mm}^2\)) and we can assume that the hole is square, we can find the initial length by taking the square root of the area: \(L = \sqrt{1.00} = 1.00 \mathrm{mm}\). Now, we can plug in the given values into the formula for linear expansion: \(\Delta L = 1.00 \mathrm{mm} \times 1.9 \times 10^{-5} \, ^{\circ} \mathrm{C}^{-1} \times 10.0^{\circ} \mathrm{C} = 1.9 \times 10^{-4} \mathrm{mm}\).

Now that we have the change in length, we can find the new length (\(L'\)) of a side of the hole by adding the change in length to the initial length: \(L' = L + \Delta L = 1.00 \mathrm{mm} + 1.9 \times 10^{-4} \mathrm{mm} = 1.00019 \mathrm{mm}\). Then, we can find the new area (\(A'\)) of the hole as the square of the new length: \(A' = (L')^2 = (1.00019 \mathrm{mm})^2 = 1.00038 \mathrm{mm}^2\).

Finally, we can find the difference in area by subtracting the initial area from the new area: \(\Delta A = A' - A = 1.00038 \mathrm{mm}^2 - 1.00 \mathrm{mm}^2 = 3.8 \times 10^{-4} \mathrm{mm}^2\). So, the area of the hole is \(3.8 \times 10^{-4} \mathrm{mm}^2\) larger at \(30.0^{\circ} \mathrm{C}\) compared to \(20.0^{\circ} \mathrm{C}\).